Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k^2 - 6k}{-3k^2 + 27k - 54} \div \dfrac{k^2 - 4k}{k^2 - 8k + 16} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{k^2 - 6k}{-3k^2 + 27k - 54} \times \dfrac{k^2 - 8k + 16}{k^2 - 4k} $ First factor out any common factors. $x = \dfrac{k(k - 6)}{-3(k^2 - 9k + 18)} \times \dfrac{k^2 - 8k + 16}{k(k - 4)} $ Then factor the quadratic expressions. $x = \dfrac {k(k - 6)} {-3(k - 6)(k - 3)} \times \dfrac {(k - 4)(k - 4)} {k(k - 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {k(k - 6) \times (k - 4)(k - 4) } { -3(k - 6)(k - 3) \times k(k - 4)} $ $x = \dfrac {k(k - 4)(k - 4)(k - 6)} {-3k(k - 6)(k - 3)(k - 4)} $ Notice that $(k - 6)$ and $(k - 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {k(k - 4)(k - 4)\cancel{(k - 6)}} {-3k\cancel{(k - 6)}(k - 3)(k - 4)} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $x = \dfrac {k\cancel{(k - 4)}(k - 4)\cancel{(k - 6)}} {-3k\cancel{(k - 6)}(k - 3)\cancel{(k - 4)}} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $x = \dfrac {k(k - 4)} {-3k(k - 3)} $ $ x = \dfrac{-(k - 4)}{3(k - 3)}; k \neq 6; k \neq 4 $